Solution Found!
Assume that the weight X in ounces of a “10- ounce” box of
Chapter 8, Problem 2E(choose chapter or problem)
Assume that the weight \(X\) in ounces of a "10ounce" box of cornflakes is \(N(\mu, 0.03)\). Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from this distribution.
(a) To test the hypothesis \(H_{0}: \mu \geq 10.35\) against the alternative hypothesis \(H_{1}: \mu<10.35\), what is the critical region of size \(\alpha=0.05\) specified by the likelihood ratio test criterion? HinT: Note that if \9\mu \geq 10.35\) and \(\bar{x}<10.35\), then \(\widehat{\mu}=10.35\).
(b) If a random sample of \(n=50\) boxes yielded a sample mean of \(bar{x}=10.31\), is \(H_{0}\) rejected? Hint: Find the critical value \(z_{\alpha}\) when \(H_{0}\) is true by taking \(\mu=10.35\), which is the extreme value in \(\mu \geq 10.35\).
(c) What is the \(p\)-value of this test?
Equation Transcription:
.
Text Transcription:
X
N(mu,0.03)
X_1,X_2,,X_n
H_0:10.35
H_1:<10.35
alpha=0.05
mu > or = 10.35
Bar x <10.35
Wide hat mu=10.35
n=50
Bar x=10.31
H_0
z_alpha
mu =10.35
mu > or = 10.35 .
p
Questions & Answers
QUESTION:
Assume that the weight \(X\) in ounces of a "10ounce" box of cornflakes is \(N(\mu, 0.03)\). Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from this distribution.
(a) To test the hypothesis \(H_{0}: \mu \geq 10.35\) against the alternative hypothesis \(H_{1}: \mu<10.35\), what is the critical region of size \(\alpha=0.05\) specified by the likelihood ratio test criterion? HinT: Note that if \9\mu \geq 10.35\) and \(\bar{x}<10.35\), then \(\widehat{\mu}=10.35\).
(b) If a random sample of \(n=50\) boxes yielded a sample mean of \(bar{x}=10.31\), is \(H_{0}\) rejected? Hint: Find the critical value \(z_{\alpha}\) when \(H_{0}\) is true by taking \(\mu=10.35\), which is the extreme value in \(\mu \geq 10.35\).
(c) What is the \(p\)-value of this test?
Equation Transcription:
.
Text Transcription:
X
N(mu,0.03)
X_1,X_2,,X_n
H_0:10.35
H_1:<10.35
alpha=0.05
mu > or = 10.35
Bar x <10.35
Wide hat mu=10.35
n=50
Bar x=10.31
H_0
z_alpha
mu =10.35
mu > or = 10.35 .
p
ANSWER:
Step 1 of 4
The electronic configuration of chlorine is implying it has seven valence electrons and oxygen with its electronic configuration of has six valence electrons